Part A The activation energy of a certain reaction is 34.3 kJ/mol . At 30 C , th
ID: 883559 • Letter: P
Question
Part A
The activation energy of a certain reaction is 34.3 kJ/mol . At 30 C , the rate constant is 0.0170s1. At what temperature in degrees Celsius would this reaction go twice as fast?
Express your answer with the appropriate units. T2=________________
Part B
Given that the initial rate constant is 0.0170s1 at an initial temperature of 30 C , what would the rate constant be at a temperature of 160 C for the same reaction described in Part A?
Express your answer with the appropriate units. k2=__________________
Explanation / Answer
Part A
we have to use the following equation
k = k · { (Ea/R) · (1/T - 1/T) }
n(k/k) = (Ea/R) · (1/T - 1/T)
T = 1 / [ 1/T - (R/Ea)·ln(k/k) ]
make sure that temperature should be in K
and Ea should be in j/mol other wise chsnge these
= 1 / [ 1/303 K - (8.31472J/molK / 34300J)·ln(2) ]
= 319.3 K
= 319.3 - 273 = 46.2878 ºC
part B
ln(k2/0.0170 s^-1) = (34300 J/mol-K / 8.314 J/mol-K)(1/303K - 1/(433K))
ln(k2/0.0170 s^-1) = 4.1255
k2 = (0.0170 s^-1)e^(4.9204)
k2 = 1.0523 s^-1